3. Simple models - Harvard Universityacmg.seas.harvard.edu/.../lecture_eps133_chap3.pdfE X P L D t...

Preview:

Citation preview

3. Simple models

Daniel J. Jacob, Atmospheric Chemistry, Harvard University, Spring 2017

Atmospheric evolution of species X is given by the continuity equation

This equation cannot be solved exactly need to construct model (simplified representation of complex system)

[ ] ( [ ])X X X XX E X P L Dt

∂= −∇• + − −

∂U

local change in concentration

with time

transport(flux divergence;U is wind vector)

chemical production and loss(depends on concentrationsof other species)

emissiondeposition

How to design and use a model

Define problem of interest

Design model; make assumptions neededto simplify equations

and make them solvable

Evaluate model with observations

Apply model: make hypotheses, predictions

Improve model, characterize its error

Design observational system to test model

One-box model

Inflow Fin Outflow Fout

XE

EmissionDeposition

D

Chemicalproduction

P L

Chemicalloss

Atmospheric “box”;spatial distribution of X within box is not resolved

out

Atmospheric lifetime: mF L D

τ =+ + Fraction lost by export: out

out

FfF L D

=+ +

Lifetimes add in parallel: export chem dep

1 1 1 1outF L Dm m mτ τ τ τ

= + + = + +

Loss rate constants add in series:export chem dep

1k k k kτ

= = + +

Mass balance equation: sources - sinks in outdm F E P F L Ddt

= = + + − − −∑ ∑

Nitrogen dioxide (NO2) has atmospheric lifetime ~ 1 day:strong gradients away from combustion source regions

Satellite observations of NO2 columns

Carbon monoxide (CO) has atmospheric lifetime ~ 2 months:mixing around latitude bands

Satellite observations

CO2 has atmospheric lifetime ~ 100 years:global mixing

Assimilated observations

On average, only 60% of emitted CO2 remains in the atmosphere – but there is large interannual variability in this fraction

Using a box model to quantify CO2 sinksPg

C y

r-1

Special case: constant source, first-order sink

( ) (0) (1 )kt ktdm SS km m t m e edt k

− −= − ⇒ = + −

Steady state solution (dm/dt = 0)

Initial condition m(0)

Characteristic time τ = 1/k for• reaching steady state• decay of initial condition

If S, k are constant over t >> τ, then dm/dt 0 and m S/k: quasi steady state

TWO-BOX MODELdefines spatial gradient between two domains

m1 m2

F12

F21

Mass balance equations: 11 1 1 1 12 21

dm E P L D F Fdt

= + − − − +

If mass exchange between boxes is first-order:

11 1 1 1 12 1 21 2

dm E P L D k m k mdt

= + − − − +

system of two coupled ODEs (or algebraic equations if system is assumed to be at steady state)

(similar equation for dm2/dt)

Research models solve mass balance equationin 3-D assemblage of gridboxes

Solve mass balance equation for individual gridboxes

A different approach: follow air parcel moving with the wind(puff model)

CX(xo, to)

CX(x, t)

wind

In the moving puff,

XdC E P L Ddt

= + − −

…no transport terms! (they’re implicit in the trajectory)

We can also have the puff dilute with background air (plume model)

CX

CX,bbackground air

,( )Xdilution X X b

dC E P L D k C Cdt

= + − − − −In plume,

Application: transport and evolution of fire plumes

Fire plumes oversouthern California

Sulfur dioxide (SO2) from Aleutian volcano eruption (8/8/2008)

http://www.avo.alaska.edu/

Kasatochi volcano

Altitude: 314 mLatitude: 52.16°NLongitude: 175.51° W

Recommended